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Suppose we have $u \in L^2(0,T;H^1(\Omega))$, and $v \in L^2(0,T;H^{-1}(\Omega))$ is the weak time derivative of $u$, so by definition it satisfies $$\int_0^T u(t)\phi'(t) = -\int_0^T v(t)\phi(t)$$ for all $\phi \in C_c^\infty(0,T)$.

My question is how to interpret the RHS. Should I think of $v(t)\phi(t)$ as $v(t)(\phi(t)$) (as $v(t) \in H^{-1}$)?

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by definition, we have $v(t)\in H^{-1}(\Omega)$ for every $t$ and it satisfies some integrability condition. So, $v(t)$ acts on the space $H^1$, of which $\phi$ is clearly a member. Therefore you interpretation was absolutely right, that $v(t)\phi(t)$ is actually $v(t)(\phi(t))$ –  math Dec 4 '12 at 13:45

2 Answers 2

The duality is misplaced. If you have a function from $(0,T)$ to a vector space, its derivative, strong or weak, is also a (generalized) function from $(0,T)$ to the same vector space. This is in fact clear from the definition you give of a weak derivative. Here, both $u$ and $u'$ are in $L^2((0,T), H^1(\Omega))$. On the other hand, if $u \in H^1((0,T), H^1(\Omega))$, then $u' \in H^{-1}((0,T), H^1(\Omega))$.

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Thanks. What is definition of $H^{-1}((0,T),H^1(\Omega)).$ Never seen that space before. –  maximumtag Apr 28 '13 at 21:00

Both the comment in the OP and the answer by Deane are wrong. $u'$ certainly need not be in $L^2(0,T;H^1)$.

You should think of $v(t)\phi(t)$ as simply multiplcation: the integral equation you wrote makes sense because the LHS is in $H^1$ and the RHS is in $H^{-1}$, and $H^1 \subset H^{-1}$.

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